In February, Michaela is going to Dubai!!
She is going to Dubai because she is smarter than all of us put together. :)
My little 13 year old daughter will be going with two other middle school kids from the school to participate in a Hungarian Math Competition called KOMAL ...and it's insane. I can't even begin to figure out the answers ...I get half way through the questions and I'm completely lost.
She has been studying with "her math team" once a week during lunch, and Happy helps her at home when necessary. She is thoroughly annoyed with me every time she asks me a question, because, although sometimes the question looks familiar, I am never quite sure on exactly what to do. Good thing Happy is smart.
I have cut and pasted some of the questions that her math coach gave her to practice on. Let me know if you figure out any of the answers!!
KÖMAL Questions:
If 73 hens lay 73 dozen eggs in 73 days and 37 hens eat 37 kg of wheat in 37 days, then how many kg of wheat are needed to produce a dozen eggs?
If the last digits of the products 1.2, 2.3, 3.4, ..., n(n+1) are added, the result is 2010. How many products are used?
Solve the following equations, where x and y denote positive prime numbers. a) xy(x+y)=2010, b) xy(x+y)=2009.
The sum of six consecutive integers is multiplied by the sum of the next six integers. Prove that the product obtained in this way will always leave the same remainder when divided by 36.
She is going to Dubai because she is smarter than all of us put together. :)
My little 13 year old daughter will be going with two other middle school kids from the school to participate in a Hungarian Math Competition called KOMAL ...and it's insane. I can't even begin to figure out the answers ...I get half way through the questions and I'm completely lost.
She has been studying with "her math team" once a week during lunch, and Happy helps her at home when necessary. She is thoroughly annoyed with me every time she asks me a question, because, although sometimes the question looks familiar, I am never quite sure on exactly what to do. Good thing Happy is smart.
I have cut and pasted some of the questions that her math coach gave her to practice on. Let me know if you figure out any of the answers!!
KÖMAL Questions:
If 73 hens lay 73 dozen eggs in 73 days and 37 hens eat 37 kg of wheat in 37 days, then how many kg of wheat are needed to produce a dozen eggs?
If the last digits of the products 1.2, 2.3, 3.4, ..., n(n+1) are added, the result is 2010. How many products are used?
Solve the following equations, where x and y denote positive prime numbers. a) xy(x+y)=2010, b) xy(x+y)=2009.
The sum of six consecutive integers is multiplied by the sum of the next six integers. Prove that the product obtained in this way will always leave the same remainder when divided by 36.
7 comments:
Um, no thank you. After the first two questions, I could barely coherently read the rest of them. I could use the excuse that kids sucked all my brains out, but I'm not sure I could have even figured those out the last time I took a math class which was in 1997.
ACTUALLY the ones we practice with a a LOT easier. But they're still really weird. There was one, I remember, it was like which one of these ropes will tangle and had four pictures of ropes... it was weird...
I'm pretty sure the real answer is: "Twice as much wheat as before is needed, because Greece and Italy is taking half of what Germany was providing for their chickens". Or I could just guess 2kg.
The first one is 73/37, about 1.97 kg
One hen lays one dozen eggs in 73 days; one hen eats 1kg in 37 days; so the answer is 73/37 kg.
The product's final digits repeat every fifth product as 2,6,2,0,0 (sum = 10) so 2010 is 201 products minus the two final zeroes we didn't need => 199
2010 = 2x3x5x67 so there's no solution
2009 = 7x7x41 so there's no solution
n+...(n+5) = 6n + 15
(n+6)+...(n+11) = 6n + 15 + 36
working mod 36, the +36 is the same as +0 and disappears.
(6n + 15)^2 = 36(n^2 + 5n) + C
working mod 36, 36(...) = 0(...) and disppears.
leaving a constant mod 36 which is a constant remainder independent of n.
--
Roger (Alex's dad)
Huh, got this wrong...
The product's final digits repeat every fifth product as 2,6,2,0,0 (sum = 10) so 2010 is 201 products minus the two final zeroes we didn't need => 199
... should be 201 sets of five products minus the final two zeroes => 1003
I'm so confused... Roger, I'm impressed you got it right, but I still don't even understand the answer! :) Better Michaela than me!!
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